导入角度js时出错答案

【问题标题】：error in importing angular js导入角度js时出错
【发布时间】：2018-02-06 18:35:18
【问题描述】：

<!DOCTYPE html>
<html>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"></script>
<body>

  <div ng-app="myApp" ng-controller="myCntrl">
    First Name : <input type="text" ng-model="firstName"> <br>
    Last Name : <input type="text" ng-model="lastName"> <br>
    <br>
    Full Name: {{firstName + " " + lastName}}
  </div>

  <script>
  var person={
    firstName = "john" ,
    lastName = "Doe"
  };

  var app = angular.module('myApp', []);
  app.controller('myCntrl', function($scope){
    $scope.firstName = person.firstName;
    $scope.lastName = person.lastName
  });
  </script>
</body>
</html>

但我在运行时得到了这个

angular.js:38 未捕获的错误：[$injector:modulerr] http://errors.angularjs.org/1.6.4/$injector/modulerr?p0=myApp&p1=Error%3A%20%5B%24injector%3Anomod%5D%20http%3A%2F%2Ferrors .angularjs.org%2F1.6.4%2F%24injector%2Fnomod%3Fp0%3DmyApp%0A%20%20%20%20at%20https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4 %2Fangular.min.js%3A6%3A425%0A%20%20%20%20at%20https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4%2Fangular.min.js%3A26 %3A270%0A%20%20%20%20at%20b%20(https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4%2Fangular.min.js%3A25%3A299)% 0A%20%20%20%20at%20https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4%2Fangular.min.js%3A26%3A44%0A%20%20%20% 20at%20https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4%2Fangular.min.js%3A42%3A117%0A%20%20%20%20at%20q%20(https% 3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4%2Fangular.min.js%3A7%3A495)%0A%20%20%20%20at%20g%20(https%3A%2F %2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6 .4%2Fangular.min.js%3A41%3A476)%0A%20%20%20%20at%20eb%20(https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4% 2Fangular.min.js%3A46%3A44)%0A%20%20%20%20at%20c%20(https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4%2Fangular.min .js%3A21%3A373)%0A%20%20%20%20at%20Sc%20(https%3A%2F%2Fajax.googleapis.com%2Fajax%2Flibs%2Fangularjs%2F1.6.4%2Fangular.min.js% 3A22%3A179) 在 angular.js:38 在 angular.js:4920 在 q (angular.js:403) 在 g (angular.js:4880) 在 eb (angular.js:4802) 在 c (angular.js:1914) 在 Sc (angular.js:1935) 在 ue (angular.js:1820) 在 angular.js:33367 在 HTMLDocument.b (angular.js:3431)

有什么帮助吗？

【问题讨论】：

标签： javascript angularjs

【解决方案1】：

您必须在 angular.js 范围内使用正确的语法正确声明 person 对象。所以将你的对象移动到控制器内。

  var app = angular.module('myApp', []);
  app.controller('myCntrl', function($scope) {
      var person = {
          firstName: "john",
          lastName: "Doe"
      };
      $scope.firstName = person.firstName;
      $scope.lastName = person.lastName
  });

<!DOCTYPE html>
<html>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"></script>

<body>
    <div ng-app="myApp" ng-controller="myCntrl"> First Name :
        <input type="text" ng-model="firstName">
        <br> Last Name :
        <input type="text" ng-model="lastName">
        <br>
        <br> Full Name: {{firstName + " " + lastName}} </div>
</body>

</html>

【讨论】：

出了什么问题？
检查更新的答案。您的对象声明不正确，最好将 person 用作 angularjs 应用程序范围内的对象，而不是将其声明为全局对象。
即使你全局声明变量，它也可以工作，虽然不推荐